# Finding the area of convergence and sum function of complex series

I am looking for a general method to find the area of convergence and sum function of the following complex series: $$\sum_{n \geq 0} \frac{2^n z^{1+n}}{(2n)!} + \sum_{n<0} \frac{(-1)^n z^{3n}}{(n-1)}.$$ and $$\sum_{n=1}^\infty n^2 z^{3n} + \sum_{n=0}^\infty \frac{z^{-n}}{(2n)!}.$$ For the first one, my attempt so far: $$\sum_{n \geq 0} \frac{2^n z^{1+n}}{(2n)!} = z \sum_{n \geq 0} \frac{(2z)^n}{(2n)!} = \sum_{n \geq 0} \frac{(\sqrt{2z})^{2n}}{(2n)!} = z \cosh(\sqrt{2}\sqrt{z}),$$ which converges everywhere on $$\mathbb{C}$$, since $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$. I'm also trying to use the Cauchy product on some known sequences like $$e^z, \sin z, \cos z$$, etc, but it doesn't seem to work so far, and it seems rather ad hoc and guessing. Is there a more general method that can be used to solve these?

The second sum change the sign of $$n$$, it is almost the sum for $$\ln(1 + x^3)$$. The third you'll get from the geometric series by taking derivative, then multiplying by $$x$$, twice (one round gives a factor $$n$$), the fourth you get from $$\cos$$.
• Making it positive gives $\sum_{n > 0} \frac{(-1)^{n+1} z^{-3n}}{n+1}$, is this correct? What is the sum for $\ln x^3$? I have not seen that as a sum series. – Sigurd Mar 4 '20 at 20:49
• @Sigurd, for $\ln(1 + x^3)$, replace $x$ by $x^3$ – vonbrand Mar 5 '20 at 21:25